Kisi bhi example se pehle, hume pin karna hoga ki "+θ" ka matlab kya hai — warna neeche har sign free float karta rahega.
Neeche ka intro figure axis-3 case draw karta hai taaki tum use dekh sako.
Figure padhna (ye memorise karo — har example isi par lean karta hai).Blue pair original axes hain: a^1 right point karta hai, a^2 upar. Green pair axis 3 ke baare mein positive yaw ke baad naye axes hain: ye blue axes hain jo counter-clockwise +θ se rotate hue hain (yellow arc +θ mark karta hai, a^1→a^2 sweep karta hua, exactly axis 3 ke liye right-hand rule). White arrow physical vector hai — wo kabhi move nahi karta. Kyunki axes +θ se spin hui lekin vector wahi raha, green frame mein vector ke coordinates aise change hote hain jaise vector −θ se spin hua ho. Wo single asymmetry hi har minus sign ka source hai jo tum aage milne wale ho.
Is topic mein jo bhi problem aa sakti hai wo in cells mein se kisi ek mein land hoti hai. Neeche ke examples mein se har ek us cell ke saath tagged hai jise wo cover karta hai.
Cell
Kya mushkil banata hai
Covered by
C1 Single rotation, angle quadrant I mein (0<θ<90°)
baseline, dono sin & cos positive
Ex 1
C2 Single rotation, angle quadrants II–IV mein (sign flips)
sin,cos sign change karte hain — positive assume mat karo
Ex 2 (saare teen quadrants)
C3 Zero / degenerate input (θ=0 ya vector 0 hai)
matrix identity ban jaata hai; "kuch nahi hota" pehchanno
Ex 3
C4 Limiting value (θ→90°,180°)
components swap ya cleanly negate hote hain — ek sanity anchor
Ex 4
C5 Which-axis minus-sign trap (C2 vs C1,C3)
wo akela matrix jisme −sinupper off-diagonal par hai
Ex 5
C6 Chained rotations + direction (CIB vs CBI)
order matters, aur inner index data se match karna chahiye
Forecast: length 62+82=10 hai. Guess karo: kya vB ki length abhi bhi 10 hogi? Kaun sa component badhega, kaun sa ghategaa?
Yaw matrix likho.Ye step kyun? Yaw axis 3 ke baare mein rotation hai, isliye hum exactly C3 use karte hain jaisa Cij=b^i⋅a^j se derive kiya gaya:
C3(40°)=cos40°−sin40°0sin40°cos40°0001=0.766−0.64300.6430.7660001
Data par apply karo.Ye step kyun? Hamara data A mein hai, aur CBA=C3(40°) ek A-vector khata hai — inner index match karta hai, isliye hum seedha multiply kar sakte hain:
vB=0.766−0.64300.6430.7660001680=0.766(6)+0.643(8)−0.643(6)+0.766(8)0=9.742.270 m/s
Result padho. Yahan ke saare angles quadrant I mein hain isliye dono sin aur cos positive hain — koi sign surprise nahi. Intro figure se connect karo: axes +40° swing huin, isliye unke relative vector second axis mein thoda kam lean karta hai — isliye 2.27<8 hai.
Verify: length 9.742+2.272=94.87+5.15=100.02≈10.0 m/s ✓ (orthogonality magnitude preserve karta hai). Units: m/s in, m/s out ✓.
Forecast: har quadrant mein ek alag (sign of cos,sign of sin) milega. Computing se pehle ye fill karo:
QII (−,+), QIII (−,−), QIV (+,−). Pehla output component 10cosψ hai, doosra −10sinψ hai.
Quadrant II, ψ=150°.Ye step kyun? Yahan cos150°=−0.866 (negative) lekin sin150°=+0.5 (positive) hai — ek mixed-sign case jise tum shortcut nahi kar sakte:
\mathbf v^B=\begin{pmatrix}-8.66\\ -5.0\\0\end{pmatrix}\text{ m/s}.$$
Doosra component $-10\sin150°=-5$ hai: negative hai even though $\sin$ positive tha, kyunki formula ka leading minus hai.
2. **Quadrant III, $\psi=210°$.** *Ye step kyun?* Ab **dono** negative hain, $\cos210°=-0.866,\ \sin210°=-0.5$:
$$\mathbf C_3(210°)=\begin{pmatrix}-0.866 & -0.5 & 0\\ 0.5 & -0.866 & 0\\ 0&0&1\end{pmatrix},\qquad
\mathbf v^B=\begin{pmatrix}-8.66\\ 5.0\\0\end{pmatrix}\text{ m/s}.$$
Yahan doosra component *positive* flip ho jaata hai ($-10\sin210°=+5$) — case (a) ka opposite, ye prove karta hai ki sign sach mein quadrant par depend karta hai.
3. **Quadrant IV, $\psi=300°$.** *Ye step kyun?* Ab $\cos300°=+0.5$ (positive) aur $\sin300°=-0.866$ (negative) hai, aakhri sign combination:
$$\mathbf C_3(300°)=\begin{pmatrix}0.5 & -0.866 & 0\\ 0.866 & 0.5 & 0\\ 0&0&1\end{pmatrix},\qquad
\mathbf v^B=\begin{pmatrix}5.0\\ 8.66\\0\end{pmatrix}\text{ m/s}.$$
Pehla component wapas positive, doosra ab positive bhi. Intro figure dekho aur imagine karo axes almost ek poora chakkar swing hui — fixed vector naye second axis mein strongly lean karta hua end hota hai.
> **Verify:** har case mein: $\sqrt{8.66^2+5^2}=10$ (QII, QIII) aur $\sqrt{5^2+8.66^2}=10$ (QIV) m/s ✓. Saare char sign combinations $(-,+),(-,-),(+,-)$ aur Example 1 ka QI $(+,+)$ ab cover ho gaye hain.
---
## Example 3 — degenerate / zero inputs (Cell C3)
> [!example]
> (a) Ek vector $\mathbf v^A=(3,-4,12)$ m/s ko **zero** rotation $\psi=0°$ se transform kiya jaata hai. (b) Zero vector $\mathbf v^A=(0,0,0)$ ko kisi bhi DCM se transform kiya jaata hai. Dono results do aur explain karo.
**Forecast:** ek rotation "kuch nahi" se kya hona chahiye? Koi bhi rotation length zero ke vector ke saath kya karna chahiye?
1. **Case (a): $\theta=0$ identity deta hai.** *Ye step kyun?* $\theta=0$ plug karo: $\cos0=1$, $\sin0=0$, toh
$$\mathbf C_3(0°)=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=\mathbf I,\qquad \mathbf v^B=\mathbf I\,\mathbf v^A=(3,-4,12)\text{ m/s}.$$
Numbers unchanged hain — axes aur vector ki description identical hain kyunki humne kuch twist nahi kiya. Intro figure mein, ye wo moment hai jab blue aur green arrows coincide hote hain. Ye *degenerate baseline* hai: har DCM $\mathbf I$ mein reduce ho jaata hai jab uska angle zero ho.
2. **Case (b): zero vector ek fixed point hai.** *Ye step kyun?* Kisi bhi matrix $\mathbf C$ ke liye, $\mathbf C\,\vec 0=\vec 0$. Length zero ke vector ki re-express karne ke liye koi direction nahi hai, isliye **har** frame agree karta hai ki wo $(0,0,0)$ hai.
> **Verify:** (a) $\mathbf I\,\mathbf v=\mathbf v$ $(3,-4,12)$ ko untouched chhodta hai, length $\sqrt{9+16+144}=13$ pehle aur baad mein ✓. (b) length $0$ pehle aur baad mein ✓.
---
## Example 4 — limiting values $\theta\to90°,180°$ (Cell C4)
> [!example]
> $\mathbf v^A=(2,7,0)$ lo aur $\mathbf C_3$ se transform karo (a) $\theta=90°$ aur (b) $\theta=180°$ par. Pehle predict karo, phir compute karo.
**Forecast:** $90°$ par axes roles swap karte hain; $180°$ par wo opposite point karte hain. Guess karo $(2,7)$ kaise move karta hai.
1. **$\theta=90°$: axis swap.** *Ye step kyun?* $\cos90°=0,\ \sin90°=1$, toh
$$\mathbf C_3(90°)=\begin{pmatrix}0&1&0\\-1&0&0\\0&0&1\end{pmatrix},\qquad \mathbf v^B=\begin{pmatrix}0&1&0\\-1&0&0\\0&0&1\end{pmatrix}\begin{pmatrix}2\\7\\0\end{pmatrix}=\begin{pmatrix}7\\-2\\0\end{pmatrix}.$$
Old $\hat a_2$-component ($7$) new first component ban jaata hai, aur old first ($2$) negated second ban jaata hai — ek clean $90°$ shuffle. Intro figure mein, green axes ko exactly quarter-turn par swing karo: naya first axis wahan land karta hai jahan old second wala point kar raha tha. Ye ek memorable **sanity anchor** hai: jab bhi tumhe sign par doubt ho, apna matrix $90°$ par test karo.
2. **$\theta=180°$: negation.** *Ye step kyun?* $\cos180°=-1,\ \sin180°=0$, toh $\mathbf C_3(180°)=\mathrm{diag}(-1,-1,1)$ aur
$$\mathbf v^B=(-2,-7,0).$$
In-plane components simply sign flip karte hain kyunki axes ab backwards point kar rahe hain, jabki axis-3 component untouched hai (hum ne uske *baare mein* rotate kiya tha).
> **Verify:** (a) $\sqrt{7^2+2^2}=\sqrt{53}$, same as $\sqrt{2^2+7^2}$ ✓. (b) $\sqrt{(-2)^2+(-7)^2}=\sqrt{53}$ ✓. Dono length preserve karte hain.
---
## Example 5 — $\mathbf C_2$ minus-sign trap (Cell C5)
> [!example]
> $\theta=30°$ ka ek pitch rotation axis 2 ke baare mein $\mathbf v^A=(1,0,0)$ m/s par act karta hai. $\mathbf v^B$ nikalo aur confirm karo ki minus sign wahan hai jahan hona chahiye.
**Forecast:** $\mathbf C_1$ aur $\mathbf C_3$ $-\sin$ diagonal ke *neeche* rakhte hain. Kya $\mathbf C_2$ bhi aisa hi karega? (Nahi — ye trap hai.)
> [!intuition] Kyun $\mathbf C_2$ ka minus *diagonal ke upar* baithta hai — cyclic order concrete bana ke
> Right-hand-rule cyclic order $1\to2\to3\to1$ yaad karo: har positive rotation ek axis ko is loop mein **agle** axis ki taraf le jaata hai. Axis 3 ke liye, positive $\theta$ $\hat a_1\to\hat a_2$ bhejtai hai ("$1$ phir $2$" direction — *increasing* index), isliye $+\sin$ upper-right mein appear karta hai aur $-\sin$ *neeche* land karta hai. Axis 2 ke liye, positive $\theta$ $\hat a_3\to\hat a_1$ bhejtai hai — wo index $3$ se $1$ tak "wrap around" karta hai, matlab wo $(1,3)$ pair of slots mein *opposite* tarike se run karta hai. Wo reversal exactly hi minus sign ko **upper** off-diagonal par flip karta hai. Axis 1 ($\hat a_2\to\hat a_3$, increasing) axis 3 ki tarah behave karta hai. Toh $\mathbf C_2$ odd one out hai purely isliye kyunki uska cyclic pair $(3\to1)$ matrix indices mein *backwards* count karta hai.
1. **$\mathbf C_2$ definition se likho, memory se nahi.** *Ye step kyun?* Kyunki axis-2 rotation cyclic order $3\to1$ follow karta hai (upar explain kiya), uska $-\sin$ **upper** off-diagonal par land karta hai (teen matrices intro formula box mein compare karo):
$$\mathbf C_2(30°)=\begin{pmatrix}\cos30° & 0 & -\sin30°\\ 0&1&0\\ \sin30° & 0 & \cos30°\end{pmatrix}=\begin{pmatrix}0.866 & 0 & -0.5\\ 0&1&0\\ 0.5 & 0 & 0.866\end{pmatrix}$$
2. **Apply karo.** *Ye step kyun?* Data ek $A$-vector hai aur $\mathbf C_2(30°)$ $A\to B$ map karta hai, toh seedha multiply karo:
$$\mathbf v^B=\mathbf C_2(30°)\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}0.866\\0\\0.5\end{pmatrix}\text{ m/s}$$
Old axis-1 ke along vector ek **positive** axis-3 component ($+0.5$) pick up karta hai, jo physical intuition se match karta hai: axes ko neeche pitch karna ek baar-horizontal vector ko tilt karta hai taaki wo new frame mein partly "up" read kare.
> **Verify:** $\sqrt{0.866^2+0.5^2}=\sqrt{0.75+0.25}=1$ m/s ✓. $-0.5$ entry is vector ko kabhi touch nahi ki (ye zero axis-3 input ko multiply karta hai), lekin ye kisi bhi axis-3 component ka sign flip *kar deta* — prove karta hai ki minus upper slot mein rehta hai.
---
## Example 6 — chained rotations & direction (Cell C6)
> [!example]
> Ek rocket ka attitude yaw $\psi=90°$ phir pitch $\theta=30°$ hai, toh $\mathbf C^{BI}=\mathbf C_2(30°)\mathbf C_3(90°)$. **Inertial** frame mein measure ki gayi ek velocity $\mathbf v^I=(0,100,0)$ m/s hai. Use **body** frame mein $\mathbf v^B$ nikalo.
**Forecast:** data inertial hai aur hum body chahte hain — kya hum $\mathbf C^{BI}$ as-is use karte hain, ya uska transpose?
1. **Inner index match karo.** *Ye step kyun?* $\mathbf C^{BI}$ *output $B$ ← input $I$* read hota hai — ye ek inertial vector khata hai aur body vector return karta hai. Hamara data inertial hai, isliye hum $\mathbf C^{BI}$ **directly** apply karte hain; koi transpose needed nahi.
2. **Composite banao (pehla rotation right par).** *Ye step kyun?* Yaw physically pehle hota hai, isliye $\mathbf C_3(90°)$ vector ko pehle touch karna chahiye — matlab wo product mein **rightmost** baithta hai; use left par rakhna galat order mein rotate karna hoga kyunki rotations commute nahi karte:
$$\mathbf C_3(90°)=\begin{pmatrix}0&1&0\\-1&0&0\\0&0&1\end{pmatrix},\quad \mathbf C_2(30°)=\begin{pmatrix}0.866&0&-0.5\\0&1&0\\0.5&0&0.866\end{pmatrix}$$
$$\mathbf C^{BI}=\mathbf C_2(30°)\mathbf C_3(90°)=\begin{pmatrix}0&0.866&-0.5\\-1&0&0\\0&0.5&0.866\end{pmatrix}$$
3. **Apply karo.** *Ye step kyun?* Composite ko inertial data mein multiply karo taaki usi physical velocity ko body axes mein read karo:
$$\mathbf v^B=\mathbf C^{BI}\begin{pmatrix}0\\100\\0\end{pmatrix}=\begin{pmatrix}0(0)+0.866(100)-0.5(0)\\ -1(0)+0(100)+0\\ 0(0)+0.5(100)+0.866(0)\end{pmatrix}=\begin{pmatrix}86.6\\0\\50.0\end{pmatrix}\text{ m/s}$$
> **Verify:** $\sqrt{86.6^2+50^2}=\sqrt{7499.6+2500}=\sqrt{9999.6}\approx100$ m/s ✓. Order matters: agar hum $\mathbf C_3(90°)\mathbf C_2(30°)$ use karte toh ek alag — galat — answer milta, kyunki rotations commute nahi karte.
---
## Example 7 — real-world thrust vectoring (Cell C7)
> [!intuition] Key idea jo tumhe [[Thrust vectoring and force resolution]] se chahiye
> Jab ek engine gimbal karta hai, thrust *body frame mein* ek fixed vector hota hai (tum uske body components gimbal angle se jaante ho). Ye find karne ke liye ki wo rocket ko sky mein kaise accelerate karta hai, tumhe usi arrow ko **inertial** frame mein re-express karna hoga — ek pure DCM job. Deflection angle aur attitude angle jab ek rotation axis share karte hain toh simply **add** ho jaate hain, jo neeche sanity check hai.
> [!example]
> Ek rocket ke engine ka gimbal aise hai ki uska thrust body direction $\hat u^B=(\cos5°,\,\sin5°,\,0)$ mein $80{,}000$ N magnitude ke saath point karta hai (body axis se ek chhota $5°$ deflection). Rocket ka ground ke relative attitude ek pure yaw $\psi=60°$ hai: $\mathbf C^{BI}=\mathbf C_3(60°)$. **Inertial** (ground) frame mein thrust vector kya hai?
**Forecast:** total ground thrust magnitude $80{,}000$ N rehni chahiye. Ground direction roughly $60°+5°=65°$ inertial-$x$ se off honi chahiye — end mein check karo.
1. **Body thrust vector form karo.** *Ye step kyun?* Magnitude times unit direction body components deta hai:
$$\mathbf F^B=80000\,(\cos5°,\sin5°,0)=(79695.6,\,6972.5,\,0)\text{ N}$$
2. **Sahi transform choose karo.** *Ye step kyun?* Hamare paas body vector hai, inertial chahiye: $\mathbf F^I=\mathbf C^{IB}\mathbf F^B=(\mathbf C^{BI})^\top\mathbf F^B$, kyunki $\mathbf C^{BI}$ inertial→body map karta hai aur humhe reverse chahiye, jo ek orthogonal DCM ke liye simply transpose hai. Kyunki $\mathbf C^{BI}=\mathbf C_3(60°)$,
$$(\mathbf C^{BI})^\top=\mathbf C_3(60°)^\top=\begin{pmatrix}\cos60° & -\sin60° & 0\\ \sin60° & \cos60° & 0\\0&0&1\end{pmatrix}=\begin{pmatrix}0.5 & -0.866 & 0\\ 0.866 & 0.5 & 0\\0&0&1\end{pmatrix}$$
3. **Transform apply karo.** *Ye step kyun?* $(\mathbf C^{BI})^\top$ ko body thrust mein multiply karna usi physical force ko ground axes mein re-express karta hai — ye question ka actual answer hai:
$$\mathbf F^I=\begin{pmatrix}0.5 & -0.866 & 0\\ 0.866 & 0.5 & 0\\0&0&1\end{pmatrix}\begin{pmatrix}79695.6\\6972.5\\0\end{pmatrix}=\begin{pmatrix}0.5(79695.6)-0.866(6972.5)\\ 0.866(79695.6)+0.5(6972.5)\\0\end{pmatrix}=\begin{pmatrix}33809.7\\72499.8\\0\end{pmatrix}\text{ N}$$
> **Verify:** magnitude $\sqrt{33809.7^2+72499.8^2}\approx80000$ N ✓. Ground direction angle $\arctan(72499.8/33809.7)=65.0°=60°+5°$ ✓ — yaw aur gimbal angles add ho gaye, exactly jaisa intuition box ne predict kiya tha.
---
## Example 8 — exam twist: angle recover karo (Cell C8)
> [!example]
> Tumhe ek DCM diya gaya hai jo axis 3 ke baare mein single rotation hai:
> $$\mathbf C_3(\theta)=\begin{pmatrix}0.6 & 0.8 & 0\\ -0.8 & 0.6 & 0\\0&0&1\end{pmatrix}$$
> Rotation angle $\theta$ nikalo. (Ye machine ko **backwards** chalata hai.)
**Forecast:** top-left $\cos\theta=0.6$ hai aur top-right $\sin\theta=0.8$ hai. Kaun sa quadrant? Dono positive → quadrant I.
1. **Dono trig values read karo.** *Ye step kyun?* $\mathbf C_3$ ke liye, entry $(1,1)=\cos\theta$ aur entry $(1,2)=\sin\theta$ hai. **Dono** hona se hum quadrant pin kar sakte hain — sirf $\cos$ ya sirf $\sin$ use karne se sign ambiguity rehti hai.
2. **$\arctan$ use karo dono signs ke saath — "kaun se angle ka ye ratio hai?" ke liye tool.** *$\arctan$ kyun aur plain $\arccos$ kyun nahi?* Kyunki $\arccos$ $+\theta$ aur $-\theta$ mein fark nahi kar sakta (cosine even hai), jabki two-argument arctangent, $\operatorname{atan2}(\sin\theta,\cos\theta)$, *dono* ke signs use karke angle ko correct quadrant mein place karta hai:
$$\theta=\operatorname{atan2}(0.8,\,0.6)=53.13°$$
3. **Recognisable triple se cross-check karo.** *Ye step kyun?* $0.6,0.8,1.0$ $3\text{-}4\text{-}5$ triangle hai $\tfrac15$ se scale kiya hua, toh $\theta=\arctan(4/3)=53.13°$ — classic angle. Isliye exam-setters in numbers ko love karte hain.
> **Verify:** $\cos53.13°=0.6$ aur $\sin53.13°=0.8$ ✓; $\arctan(0.8/0.6)=53.13°$ ✓.
---
> [!recall]- Har nayi problem kaun se cell mein fall karegi? (saare aath cells)
> $40°$ ka ek yaw jisme dono components positive hain ::: Cell C1 (quadrant I baseline, koi sign surprise nahi).
> Ek vector par apply kiya gaya $135°$ ka yaw ::: Cell C2 (quadrant II — $\cos$ negative, $\sin$ positive).
> $\theta=0$ se transform karo ya zero vector transform karo ::: Cell C3 (identity / fixed point, kuch nahi badalta).
> Ek sign anchor ke roop mein exactly $90°$ ya $180°$ par rotation ::: Cell C4 (limiting swap / negation).
> Axis 2 ke baare mein pitch, minus sign ke baare mein worried ::: Cell C5 (upper off-diagonal, cyclic order $3\to1$ se).
> Ek body vector hai, inertial chahiye, do stacked rotations ke zariye ::: Cell C6 (transpose + correct order).
> Gimballed thrust: body force ground axes mein chahiye ::: Cell C7 (real-world resolution, angles add hote hain).
> Ek matrix diya gaya, uska angle nikalo ::: Cell C8 (atan2 se backwards chalao).
> [!mnemonic]
> **"Quadrant, zero, ninety, which-axis, which-direction, real-world, backwards"** — ye tarike hain jisme ek DCM problem bite kar sakti hai. Agar tumhara exam question kisi se match nahi karta, tumne use misread kiya hai.
Dekho bhi: [[Euler angles — yaw pitch roll]], [[Orthogonal matrices and rotation groups SO(3)]], aur moving-frame version ke liye [[Angular velocity and the kinematic equations]].
